Question: You have found the following ages (in years) of 6 turtles. Those turtles were randomly selected from the 47 turtles at your local zoo: $ 21,\enspace 84,\enspace 21,\enspace 73,\enspace 60,\enspace 43$ Based on your sample, what is the average age of the turtles? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we only have data for a small sample of the 47 turtles, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{21 + 84 + 21 + 73 + 60 + 43}{{6}} = {50.3\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {858.49} + {1135.69} + {858.49} + {515.29} + {94.09} + {53.29}} {{6 - 1}} $ {s^2} = \dfrac{{3515.34}}{{5}} = {703.07\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{703.07\text{ years}^2}} = {26.5\text{ years}} $ We can estimate that the average turtle at the zoo is 50.3 years old. There is also a standard deviation of 26.5 years.